Step by Step solution about How to resolve the algorithm Map range step by step in the Ruby programming language

The provided Ruby code defines a method called map_range that performs a linear mapping between two ranges, a and b, for a given input value s. It then demonstrates the usage of this method in two scenarios.

1. Method Definition:

def map_range(a, b, s)
 af, al, bf, bl = a.first, a.last, b.first, b.last
 bf + (s - af)*(bl - bf).quo(al - af)
end

• a: This is the first range, specified as a Ruby range or array.
• b: This is the second range, specified as a Ruby range or array.
• s: This is the input value within the first range a.

The method first extracts the first and last values of both ranges (af, al, bf, bl). It then computes the relative position of s within the first range a as a fraction using the formula (s - af) / (al - af).

It then uses this fraction to linearly interpolate between the values in the second range b using the formula bf + (bl - bf) * (fraction).

2. Usage of the Method:

(0..10).each{|s| puts "%s maps to %g" % [s, map_range(0..10, -1..0, s)]}

In this snippet, the code iterates through values s from 0 to 10. For each value of s, it calls the map_range method with the range (0..10) as the first range and (-1..0) as the second range. The result is printed, showing how each value s in the first range maps to a value in the second range. The %g format specifier is used to display the result as a floating-point number.

(0..10).each do |s|
 puts "%s maps to %s" % [s, map_range(0..10, -1..0, s)]
end

This code is similar to the first snippet but uses string interpolation to display the result as a string rather than a float.

def map_range(a, b, s)
  af, al, bf, bl = a.first, a.last, b.first, b.last
  bf + (s - af)*(bl - bf).quo(al - af)
end

(0..10).each{|s| puts "%s maps to %g" % [s, map_range(0..10, -1..0, s)]}


(0..10).each do |s|
  puts "%s maps to %s" % [s, map_range(0..10, -1..0, s)]
end